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STRUCTURE OF Mg-24 AND Mg-23
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) Despite the enormous success of the Bohr model (1913) and the quantum mechanics of Schrodinger (1926) based on the well-established laws of electromagnetism in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, after the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) neither was able to discover the correct nuclear structure, because the discovery of the assumed uncharged neutron led to the abandonment of electromagnetic laws in favour of wrong theories which could not lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for discovering the nuclear force and structure by applying the well-established laws of electromagnetism. (See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). ' Structure of Mg-24 with S = 0' Using the first diagram of the structure of Mg24 which represents the Fig.6d of my published paper one sees that here the core is not the parallelepiped of O-16 with S = 0 but a greater parallelepiped including 6 horizontal squares with S = 0. Here the nucleons p1, n1 p2, and n2 of the first horizontal square give S = -2, while the total spin of the nucleons p11, n11, p12, and n12 of the sixth horizontal square is S = +2. Since the total spin of the nucleons forming the central parallelepiped has the structure of O-16 with S = 0 one concludes that the total spin of Mg-24 is S = 0. However here the central parallelepiped with a similar structure as that of O-16 differs fundamentally from the isolated O-16 because the nucleons of all four horizontal squares as shown also in Fig 6d of my published paper have four pn bonds per nucleon. This situation contributes to the increase of the binding energy of all pn bonds which overcome the nn repulsions of short range and the pp repulsions of long range. Although the six horizontal squares lead to the anisotropy of elongated shapes the structure of Mg24 is stable because the central parallelepiped which is similar to the parallelepiped of the isolated O-16 has at all points of a symmetrical parallelepiped four pn bonds per nucleon. Under this condition the structure of Mg24 with equal number of protons and neutrons belongs to the group of the parallelepipeds called alpha particle nuclei as C-12, O-16, Ne-20 and Mg-24. Stable Mg-24 with S = 0 Unstable Mg-23 with S = +3/2 p12(+1/2.n12(+1/2) p10(+1/2)n10(+1/2) ' '''n11(+1/2).p11(+1/2) n9(+1/2).p9(+1/2) n10(-1/2).. p10(-1/2) n8(-1/2).p8(-(1/2) p9(-1/2).n9(-1/2) p7(-1/2).n7(-1/2)…p12( -1/2) p8(+1/2).n8(+1/2) p6(+1/2).n6(+1/2) n7(+1/2).p7(+1/2) n5(+1/2).p5 (+1/2)…n11(+1/2) n6(-1/2).p6(-1/2) n4(-1/2).p4(-1/2) p5 (-1/2).n5 (-1/2) p3(-1/2).n3(-1/2)…..p11( -1/2) p4(+1/2).n4(+1/2) p2(+1/2).n2(+1/2) n3(+1/2). p3(+1/2) n1(+1/2).p1(+1/2) ' n2(-1/2).p2(-1/2) p1(-1/2)n1(-1/2) ' ' '''Structure of Mg-23 . Why the unstable Mg-23 with 12 protons and 11 neutrons turns to the stable Na-23 with 11 protons and 12 neutrons In the absence of a neutron as in the case of Mg-23 with 12 protons and 11 neutrons the greater number of protons leads to the decay. Note that the proton charges of 4u = +8e/3 and 5d = -5e/3 give a net charge +e. Therefore when the number of protons is greater than the number of neutrons the repulsions could overcome the pn bonds of short range because one proton interacts with all the other protons with a repulsive force of long range. In the case of the Na23 with11 protons and 12 neutrons we discovered that it is stable because in the extra two neutrons with the one proton as n11, p11 and n12, the p11 makes the three bonds p11n6, p11n11, and p11n12 of short range able to overcome the pp repulsions of long range. Whereas the structure of Ne23 with 10 protons and 13 neutrons leads to the decay because the extra three neutrons form only single np bonds of weak binding energy. Thus the pp repulsions of long range could overcome the weak binding energy of single np bonds (See my STRUCTURE OF Na23 AND Ne23 ). However in the case of the unstable Mg23 one observes a different phenomenon due to the number of protons which overcome the number of neutrons. So in the second diagram which has the same core as that of the Na23 we observe not the two neutrons with the one proton but the two protons with the one neutron. Therefore at the second horizontal square there exist not the n11 but the p11 which makes not the three pn bonds as those of Na23 but the two bonds as p11n3 and p11n11. Moreover the extra proton p12 of the fourth horizontal square exerts on p11 an additional repulsive force . Thus the repulsions of the pp systems with a repulsive energy Q of long range could overcome the two bonds of p11. Under this condition the p11 turns into a new n as p11 + Q = n + positron + neutrino Such a new neutron ( n ) represents the n11 of Na23. In other words the three extra nucleons of the unstable Mg23 as p11, n11 and p12 are replaced by the three extra nucleons n11, p11, and n12 of the stable Na23. Category:Fundamental physics concepts